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I have recently encountered a reasoning question that I have solved half , but I can't solve one part of it. Question : \begin{align} 3 + 5 + 6 =&\; 151872 \\[1.3ex] 5 + 5 + 6 =&\; 253094 \\[1.3ex] 5 + 6 + 7 =&\; 303585 \\[1.3ex] 5 + 5 + 3 =&\; 251573 \\[1.3ex] \end{align} I could figure out that ,in the first , $1$st $*$ $2$nd ($3 * 5$) equals first two digits of the answer ($15$). This applies for all the other. Similarly , $1$st $*$ $3$rd ($3 * 6$) equals $3$rd and $4$th digit concatenated of the answer ($18$).This also applies in each. But I can't figure out how the last two digits are formed in each.

Please tell me how are the last two digits are formed.

Thanks.

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  • $\begingroup$ $102+6a-16b+7c$ works fine $\endgroup$ Aug 17, 2013 at 14:48
  • $\begingroup$ @HagenvonEitzen Please explain a bit more. Thank you. $\endgroup$ Aug 17, 2013 at 14:49
  • $\begingroup$ @HagenvonEitzen Almost. doesn't work for the first one. $\endgroup$ Aug 17, 2013 at 14:50

4 Answers 4

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If $a$ is the first number, $b$ is the second, and $c$ is the third, the required number is obtained by computing $ab + ac -c$ and then reversing the digits.

For example, in the first row:

$3\times 5 + 3 \times 6 - 6 = 27$, and then reverse to get $72$.

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  • $\begingroup$ This also seems to be the overwhelming majority opinion if one googles 151872 $\endgroup$ Oct 14, 2013 at 17:34
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Try this: $$ 11a - 16b + 7c + 77 $$ Here $a$ is the first number, $b$ the second, $c$ the third. For example $$ 11\cdot 3 -16\cdot5 + 7\cdot 6 + 77 = ... $$


How did I find this? I made the matrix $$ A = \pmatrix{5 & 5 & 6 \\ 5 & 6 & 7 \\ 5 & 5 & 3}. $$ This matrix corresponds to the last three rows above. Then we want to solve $$ A\pmatrix{x \\ y \\ z} = \pmatrix{94 \\ 85 \\ 73}. $$ And you would get $$ \pmatrix{x\\y\\z} = \pmatrix{26.4 \\ -16 \\ 7} $$ But we have $$ 26.4\cdot 3 - 16\cdot 5 + 7\cdot 6 = 41.2 \neq 72. $$ This leads to trying to add a constant to all the values of $$ \pmatrix{94 \\ 85 \\ 73} + \pmatrix{N \\ N \\ N}. $$ If you make a small program you find out that $N = - 77$ works $$ \pmatrix{17 \\ 8 \\ -4}. $$ Then $$ A^{-1}\pmatrix{17 \\ 8 \\ -4} = \pmatrix{11 \\ -16 \\ 7}. $$ So, your program should run through values of $N$ and check whether the first row is satisfied. I am not explaining this very well. And this is not a very smart way of doing it. Someone else can probably some up with something more clever.

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  • $\begingroup$ Any intuition behind this? $\endgroup$
    – Git Gud
    Aug 17, 2013 at 15:56
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It can be represented as some "positional notation":

$ {\Large 3}\cdot 50611 + {\Large 5} \cdot 49984 + {\Large 6} \cdot 507 - 252923 = {\Large 151872}$,

$ {\Large 5}\cdot 50611 + {\Large 5} \cdot 49984 + {\Large 6} \cdot 507 - 252923 = {\Large 253094}$,

$ {\Large 5}\cdot 50611 + {\Large 6} \cdot 49984 + {\Large 7} \cdot 507 - 252923 = {\Large 303585}$,

$ {\Large 5}\cdot 50611 + {\Large 5} \cdot 49984 + {\Large 3} \cdot 507 - 252923 = {\Large 251573}$.

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  • $\begingroup$ Could you explain a bit little more . Thanks. $\endgroup$ Aug 18, 2013 at 3:45
  • $\begingroup$ @GaurangTandon, it is simply linear combination like $\begin{array}{c} 3a+5b+6c+d=151872,\\ 5a+5b+6c+d=253094,\\ 5a+6b+7c+d=303585,\\ 5a+5b+3c+d=251573;\\ \end{array}\tag{1}$ to find $a,b,c,d$, we solve linear system (1). $\endgroup$
    – Oleg567
    Aug 19, 2013 at 12:18
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    $\begingroup$ @GaurangTandon, but your observation that 3*5=15 (1st part), 3*6=18 (2nd part), and Peter Phipps' answer (3rd part) is very nice. Exactly for puzzle solving. $\endgroup$
    – Oleg567
    Aug 19, 2013 at 12:20
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ANS-366329

-36(multiplication of first two digits(9*4)) -63-(multiplication of end digits(9*7)) -29(add 36+63 and minus the last digit(7) ansd reverse the it i.e. 36+63-7=92 reverse of 92 is 29

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  • $\begingroup$ This doesn't answer the question , but well , thanks for the try, I already know the answer :) . $\endgroup$ Oct 15, 2013 at 14:09

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